Chapter 10.6, Problem 7PSA

To determine

To Find : $\stackrel{⌢}{PQ}$

Answer to Problem 7PSA

  $137°$

Explanation of Solution

Given information:

1) $\overleftrightarrow{PY}\text{ and }\overleftrightarrow{QW}\text{ are tangents to smaller circle}\text{.}$

2) $\stackrel{⌢}{WZ}=126°,\stackrel{⌢}{XY}=40°$

Formula Used:

1) the angle between two secants is equal to half the difference of the enclosed arcs.

2) The angle between two tangents is equal to half the difference of the enclosed arcs.

  $\begin{array}{l}\text{Let the intersection of tangent be called T}\text{.}\\ \frac{1}{2}(\stackrel{⌢}{WZ}-\stackrel{⌢}{XY})=\angle YTZ\\ \frac{1}{2}(126°-40°)=\angle YTX\\ \angle YTX=43°\\ \text{and }\angle PTQ=43°\because \angle YTX=\angle PTQ\\ \text{Now major }\stackrel{⌢}{PQ}+\text{minor}\stackrel{⌢}{PQ}=360°\\ \text{and }\frac{1}{2}(\text{major }\stackrel{⌢}{PQ}-\text{minor}\stackrel{⌢}{PQ})=\angle PTQ\\ \frac{1}{2}(360°-2\text{(minor }\stackrel{⌢}{PQ}\text{)})=43°\\ 360°-2(\text{minor }\stackrel{⌢}{PQ})=86°\\ 2(\text{minor }\stackrel{⌢}{PQ})=274°\\ \text{minor}\stackrel{⌢}{PQ}=137°\end{array}$